Maxwell-Boltzmann Statistics

When deriving the barometric height formula, we have already seen that the relationship between pressure and density is constant. We can therefore replace the pressure in this formula with the density of the gas without any problems: $$\varrho = \varrho_0 e^{-\frac{\varrho_0gh}{p_0}}$$ If you now expand the fraction in the exponent with an arbitrary volume $V$ and then replace the denominator with the ideal gas equation, you get $$\varrho = \varrho_0 e^{-\frac{mgh}{Nk_BT}}$$ We already know that the particle number density $n$ in a given volume is proportional to the density of the gas, i.e. it holds: $$n = n_0 e^{-\frac{mgh}{k_BT}}$$ The numerator can now be interpreted as the potential energy of the gas $E=mgh$, which makes sense since we have already seen from considerations within the kinetic theory of gases that the quantity $k_B T$ is linked to the energy of a gas. Assuming that all gas layers are in equilibrium, the law of conservation of energy can be used and the potential energy can be replaced by the kinetic energy: $$n = n_0e^{-\frac{E_\mathrm{kin}}{k_B T}}$$ The exponential term plays an important role in thermodynamics and is called Boltzmann factor. If you want to use this to calculate the probability that a particle has the velocity $v$, the resulting function still has to be normalized so that the integral under the graph has the value 1. If one now uses the formula for the kinetic energy, the following probability distribution results for a particle in a gas at temperature $T$: $$p(v) = \sqrt{\frac{m}{2\pi k_BT}}e^{-\frac{mv^2}{2k_BT}}$$ One immediately recognizes that this is a symmetrical Gaussian distribution around the origin, i.e. the mean value of all velocities is 0, since there is no preferred direction for the gas particles. For all three dimensions, the relationship $v^2 = v_x^2 + v_y^2 + v_z^2$ results in spherical surfaces whose probability density can also be described by a Gaussian distribution. With this preliminary consideration, we now want to calculate the distribution for the magnitude of the velocity without taking the direction of the particles into account: For this, we assume that all particles with the amount of velocity $v$ are on a spherical surface around the origin. For the probability of finding a particle on this spherical surface, the above-mentioned equation must therefore be multiplied by the spherical surface $4\pi v^2$ itself. The result is then the well-known Maxwell-Boltzmann distribution: $$\boxed{p(v) = \left(\sqrt{\frac{m}{2\pi k_BT}}\right)^3 4\pi v^2 e^{-\frac{mv^2}{2k_BT }}}$$ The normalization constant must also be multiplied by itself three times to get an area of ​​1, which explains the cube of the root. If you derive this function and set the result equal to 0, you get the result for the maximum of the distribution: $$\boxed{\hat{v} = \sqrt{\frac{2k_BT}{m}}}$$ This is therefore the speed with the greatest probability, but not the average speed of the gas particles. This can be calculated analogously to the arithmetic mean by summing up to overall possible speeds. Since there are infinitely many velocities, the sum becomes an integral: $$\bar{v} = \int_0^\infty vp(v)\,\mathrm{d} v$$ By partial integration one thus obtains the mean speed $$\boxed{\bar{v} = \frac{8k_BT}{\pi m}}$$ Geometrically, this is the speed at which the areas under the graph on either side of this value become equal. In addition to that, the theoretically calculated value of the mean velocity $\bar{v}$ for two different temperatures was included. One can see that the distribution function depends very strongly on the temperature $T$ and is asymmetric.